Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $22.9$ years; the standard deviation is $2.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living longer than $15.4$ years.
Solution: $22.9$ $20.4$ $25.4$ $17.9$ $27.9$ $15.4$ $30.4$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $22.9$ years. We know the standard deviation is $2.5$ years, so one standard deviation below the mean is $20.4$ years and one standard deviation above the mean is $25.4$ years. Two standard deviations below the mean is $17.9$ years and two standard deviations above the mean is $27.9$ years. Three standard deviations below the mean is $15.4$ years and three standard deviations above the mean is $30.4$ years. We are interested in the probability of a tiger living longer than $15.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $15.4$ years and the other half $({0.15\%})$ will live longer than $30.4$ years. The probability of a particular tiger living longer than $15.4$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.